Simplify the following expression: $y = \dfrac{9x^2+25x- 6}{9x - 2}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(9)}{(-6)} &=& -54 \\ {a} + {b} &=& &=& {25} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-54$ and add them together. Remember, since $-54$ is negative, one of the factors must be negative. The factors that add up to ${25}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-2}$ and ${b}$ is ${27}$ $ \begin{eqnarray} {ab} &=& ({-2})({27}) &=& -54 \\ {a} + {b} &=& {-2} + {27} &=& 25 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({9}x^2 {-2}x) + ({27}x {-6}) $ Factor out the common factors: $ x(9x - 2) + 3(9x - 2)$ Now factor out $(9x - 2)$ $ (9x - 2)(x + 3)$ The original expression can therefore be written: $ \dfrac{(9x - 2)(x + 3)}{9x - 2}$ We are dividing by $9x - 2$ , so $9x - 2 \neq 0$ Therefore, $x \neq \frac{2}{9}$ This leaves us with $x + 3; x \neq \frac{2}{9}$.